题意：

给定\(n,m,p\)，求
\[ \sum_{a=1}^n\sum_{b=1}^m\frac{\varphi(ab)}{\varphi(a)\varphi(b)}\mod p \]

思路：

由欧拉函数性质可得：\(x,y\)互质则\(\varphi(xy)=\varphi(x)\varphi(y)\)；\(p\)是质数则\(\varphi(p^a)=(p-1)^{a-1}\)。因此，由上述两条性质，我们可以吧\(a,b\)质因数分解得到
\[ \begin{aligned} \sum_{a=1}^n\sum_{b=1}^m\frac{\varphi(ab)}{\varphi(a)\varphi(b)}\mod p&=\sum_{a=1}^n\sum_{b=1}^m\frac{gcd(a,b)}{(p_1 - 1)(p_2-1)\dots (p_k-1)}\mod p\\ &=\sum_{a=1}^n\sum_{b=1}^m\frac{gcd(a,b)}{\varphi(gcd(a,b))}\mod p\\ &=\sum_{k}\sum_{k|d}\mu(\frac{d}{k})F(d)*k*inv[\varphi(k)] \mod p \end{aligned} \]
有点卡常。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1000000007;
using namespace std;int mu[maxn], vis[maxn];
int prime[maxn], cnt, phi[maxn];
ll inv[maxn];
void init(int n){for(int i = 0; i <= n; i++) vis[i] = mu[i] = 0;cnt = 0;mu[1] = 1;phi[1] = 1;for(int i = 2; i <= n; i++) {if(!vis[i]){prime[cnt++] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 0; j < cnt && prime[j] * i <= n; j++){vis[prime[j] * i] = 1;if(i % prime[j] == 0){phi[i * prime[j]] = phi[i] * prime[j];break;}mu[i * prime[j]] = -mu[i];phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}
}
void init2(int n, ll p){inv[0] = inv[1] = 1;for(int i = 2; i <= n; i++)inv[i] = (p - p / i) * inv[p % i] % p;
}int main(){init(1e6);int T;scanf("%d", &T);while(T--){ll n, m, p;scanf("%lld%lld%lld", &n, &m, &p);ll mm = min(n, m);init2(mm, p);ll ans = 0;for(int k = 1; k <= mm; k++){ll temp = 0;for(int d = k; d <= mm; d += k){temp += 1LL * mu[d / k] * (n / d) * (m / d);}temp = temp * k % p * inv[phi[k]] % p;ans = (ans + temp) % p;}ans = (ans + p) % p;printf("%lld\n", ans);}return 0;
}